Suppose that you have a supply of a 25% solution of alcohol and a 75% solution of alcohol. How many quarts of each should be mixed to produce 100 quarts that is 30% alcohol?

.25x+.75(100-x)=.30(100)

solve for x

I solved for x which is 90. what do I do next ?

thanks for your help

x=how many quart of 25% solution of alcohol

100-x=the reminder of quart; or how many quart of 75% solution of alcohol

thanks

To determine the number of quarts of each solution needed to produce a 100-quart mixture with a 30% alcohol content, we can set up a system of equations.

Let's assume x quarts of the 25% solution are mixed, and therefore, (100 - x) quarts of the 75% solution will be mixed.

The amount of alcohol in the 25% solution will be (25/100) * x = 0.25x quarts.
The amount of alcohol in the 75% solution will be (75/100) * (100 - x) = 0.75(100 - x) quarts.

Since we want the final mixture to have a 30% alcohol content, the amount of alcohol in the final mixture will be (30/100) * 100 = 30 quarts.

Now, we can write the equation for the total amount of alcohol in the mixture:
0.25x + 0.75(100 - x) = 30

Let's solve this equation to find the value of x:

0.25x + 0.75(100 - x) = 30
0.25x + 75 - 0.75x = 30
-0.5x = 30 - 75
-0.5x = -45
x = -45 / -0.5
x = 90

Therefore, you would need to mix 90 quarts of the 25% solution and (100 - 90) = 10 quarts of the 75% solution to produce 100 quarts of a 30% alcohol mixture.